Nonlinear finite elements/Homework 10/Solutions

Problem 1: Kinematics and Stress Rates edit


Figure 1 shows a linear three-noded triangular element in the reference configuration.

Figure 1. Three-noded triangular element.

The motion of the nodes is given by:


The configuration ( ) of the element at time   is given by


Solutions edit

Part 1 edit

Write down expressions for  ,  , and   in terms of the initial configuration ( ) ?

In the initial configuration  . Therefore,


Therefore, the initial configuration is given by


Substituting the values of   and  , we get




Also, the shape functions must satisfy the partition of unity condition




The required expressions are


Part 2 edit

Derive expressions for the deformation gradient and the Jacobian determinant for the element as functions of time.

The deformation gradient is given by


Before computing the derivatives, let us express   in terms of  . Recall




Substituting in the expressions for   and  , we get


In the above expression, the parent coordinates   are no longer useful. Therefore, we write


where  . Taking derivatives, we get




The Jacobian determinant is




Part 3 edit

What are the values of   and   for which the motion is isochoric?

For isochoric motion,  . Therefore,


One possibility is


This is a pure rotation.

Another possibility is that


This is a combination of shear and rotation where the volume remains constant.

Part 4 edit

For which values of   and   do we get invalid motions?

We get invalid motions when  . Let us consider the case where  . Then


This is possible when


If   then


That is,


Therefore the values at which we get invalid motions are


Part 5 edit

Derive the expression for the Green (Lagrangian) strain tensor

for the element as a function of time.

The Green strain tensor is given by




Let us make the following substitutions








The Green strain is


Part 6 edit

Derive an expression for the velocity gradient as a function of


The velocity is the material time derivative of the motion.

Recall that the motion is given by




We could compute the velocity gradient using


after expressing   in terms of  . However, that makes the expression quite complicated. Instead, we will use the relation


The time derivative of the deformation gradient is


The inverse of the deformation gradient is


Using the substitutions


we get




The product is


Note that the first matrix is symmetric while the second is skew-symmetric.

Therefore, the velocity gradient is


Part 7 edit

Compute the rate of deformation tensor and the spin tensor.

The rate of deformation is the symmetric part of the velocity gradient:


The rate of deformation is the skew-symmetric part of the velocity gradient:


Part 8 edit

Assume that   and  . Sketch the undeformed configuration and the deformed configuration at   and  .  Draw both the deformed and undeformed configurations on the same plot

and label.

Recall that in the initial configuration


Also, the motion is


Plugging in the values of   and  , we get


At  ,


At  ,


The deformed and undeformed configurations are shown below.

Deformed and undeformed configurations.

Part 9 edit

Compute the polar decomposition of the deformation gradient with the above values of   and  ,

The deformation gradient is


The right Cauchy-Green deformation tensor is


The eigenvalue problem is


This problem has a solution if




The eigenvalues are (as expected)


The principal stretches are


The principal directions are (by inspection)


Now, the right stretch tensor is given by




Hence the right stretch is


At  , we have






Therefore, the rotation is


At  , we have


Part 10 edit

Assume an isotropic, hypoelastic constitutive equation for the material of the element. Compute the material time derivative of the Cauchy stress at   using (a) the Jaumann rate and (b) the Truesdell rate.

A hypoelastic material behaves according to the relation


For an isotropic material




Recall that


Using the values of   and   from the previous part, at  ,


Therefore, the trace of the rate of defromation is




For the Jaumann rate


where the spin is






For the Truesdell rate






For  ,  ,  , we have








Problem 2: Hyperelastic Pinched Cylinder Problem edit

Read the following paper on shells:

Buchter, N., Ramm, E., and Roehl, D., 1994, "Three-dimensional extension of non-linear shell formulation based on the enhanced assumed strain concept," Int. J. Numer. Meth. Engng., 37, pp. 2551-2568.

Answer the following questions:

Solution edit

Part 1 edit

What do the authors mean by "enhanced assumed strain"?

See Wikipedia article on [w:Enhanced assumed strain|enhanced assumed strain]] .

Part 2 edit

Example 8.2 (and Figures 3 and 4 and Table III) in the paper discusses the simulation of a hyperelastic cylinder. Perform the same simulation using ANSYS for a shell thickness of 0.2 cm. Use shell elements and the Neo-Hookean hyperelastic material model that ANSYS provides.

The following material properties are used:

  kN/cm ,  ,   kN/cm , and   cm /kN. Symmetry is used and only half of the model is meshed. At the support, the model is constraint in all directions. The load of 36 kN is applied on the top of the cylinder (Fig~2. ANSYS input listing is shown Fig~6 of Appendix.

Figure 2. Meshed model

Part 3 edit

Compare the total load needed to achieve an edge displacement of 16 cm with the results given in Table III. Comment on your results.

The plot of force vs. edge displacement (vertical) and the deformed model are shown in Fig 3. From the plot, one sees that a load of 35.1 kN is required to deform the edge by 16 cm. This is less than 1% difference compared to the result given in the paper using 7-parameter shell theory.

Figure 3. Left: Force vs. edge displacement. Right: Deformed model.

Problem 3: Elastic-Plastic Punch Indentation edit

Read the following paper on elastic-viscoplastic FEA:

Rouainia, M. and Peric, D., 1998, "A computational model for elasto-viscoplastic solids at finite strain with reference to thin shell applications," Int. J. Numer. Meth. Engng., 42, pp. 289-311.

Answer the following questions:

Solution edit

Part 1 edit

Example 5.4 of the paper shows a simulation of the deformation of a thin sheet by a square punch. Perform the same simulation for 6061-T6 aluminum. Assume linear isotropic hardening and no rate dependence.

The following data are used for the thin plate:  GPa,  ,   Mpa,   Mpa, see Fig 4 for the stress-strain data. Symmetry is used and only half of the model is meshed (Fig 4). At the support, the model is constrained in all directions. A plastic finite strained shell (SHELL43) is chosen for this simulation.

The following material properties are used for the punch and die:  GPa,  . The value of Young's modulus is arbitrary chosen so long as it is high enough to remain rigid during the simulation.

The meshed model is illustrated in Fig 4. A load of 10 kN is applied on the top of the punch. Load steps are split into two steps. First load step the punch is moved close to the plate to activate the contact elements (CONTAC49). The second load step, 30 kN is applied on top of the punch. ANSYS input listing is shown Fig 7 of the Appendix.

Figure 4. Left: Bi-linear stress-strain data. Right: Meshed model.

Part 2 edit

Draw a plot of the punch force vs. punch travel and compare your result with the results shown in Figure 13 of the paper (qualitative comparison only).

The punch force vs. punch travel and the plot of the deformation at the final load step is shown in Fig 5. The curve displays similar pattern as those shown in the paper.

Figure 5. Left: Punch force vs. punch travel. Right: Sketch of deformation at final load step.

Appendix edit

Figure 6. ANSYS input listing for Problem 3.
Figure 7. ANSYS input listing for Problem 3.